0

HETAV PATEL

Grade 11,

Courses New

a certain ruby laser emits 2.5 j pulses of light of wavelenght 6940 angstrom .what is the minimum number of cromium ions in the ruby laser??

a certain ruby laser emits 2.5 j pulses of light of wavelenght 6940 angstrom .what is the minimum number of cromium ions in the ruby laser??

Grade:12th pass

2 Answers

Arun
25750 Points
6 years ago
There must be at least that many chromium 3+ ions in the crystal.
n 3.489 10 photons 18 = ×
n photons, because 1 joule per pulse divided by joules per photon gives photons per pulse. 1
E
:=
In each pulse of light, there are
E 2.866 10 Joules − 19 E = ×
h c⋅
λ
:=
The energy of a 694 nm photon is
λ 694 10− 9 c 3 10 := ⋅ 8 h 6.63 10 := ⋅ 
Emmanuel Sherman
28 Points
4 years ago
The formula says E=hc/\lambda
So let h=6.63*10e-34
     let c=3*10e8
     let \lambda=6940*10e-9m
E=hc/\lambda 
= (6.63*10e-34)(3*10e8)/6940*10e-9m
E=2.86*10e-19
Then n=1/E
=1/2.86*10e-19
n=3.496*10e18 will be the minimum number of Cr3+ in the ruby lazer
 

Think You Can Provide A Better Answer ?

Other Related Questions on Modern Physics

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More